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December 2014
Thermodynamic
Units and
Properties
Student Guide
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Table of Contents
INTRODUCTION ..................................................................................................................... 1
TLO 1 THERMODYNAMIC PROPERTIES ................................................................................. 2
Overview .......................................................................................................................... 2
ELO 1.1 Properties and Definitions ................................................................................. 3
ELO 1.2 Thermodynamic Properties of Temperature ................................................... 15
ELO 1.3 Thermodynamic Properties of Pressure .......................................................... 18
TLO 1 Summary ............................................................................................................ 24
TLO 2 CONCEPTS OF HEAT, WORK, AND ENERGY .............................................................. 25
Overview ........................................................................................................................ 25
ELO 2.1 Thermodynamic Properties of Energy ............................................................ 26
ELO 2.2 Relationship Between Work, Energy, and Power ........................................... 32
ELO 2.3 Thermodynamic Properties: Introduction to Heat........................................... 38
TLO 2 Summary ............................................................................................................ 42
THERMODYNAMIC UNITS AND PROPERTIES SUMMARY....................................................... 43
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Thermodynamic Units and Properties
Revision
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Introduction
Thermodynamics is a branch of natural science concerned with heat and its
relation to energy and work. It defines macroscopic variables (such as
temperature, internal energy, entropy, and pressure) that characterize
Rev 1
1
materials and explains the ways they are related and the laws that govern
how they change with time.
Nuclear power plants generate thermal heat energy in the reactor core and
convert it into useful mechanical work energy with a turbine. To
understand the various aspects and ramifications of this energy conversion,
we must study the interplay of three significant thermal sciences—heat
transfer, thermodynamics, and fluid flow—as they specifically relate to the
normal and abnormal operations of the plant. We must also become very
familiar with the nature and behavior of water particularly as it relates to the
pressurized water reactor (PWR) energy conversion process.
A working knowledge of thermodynamics is required for operators in the
understanding of nuclear power plants. Before studying thermodynamics, it
is important to establish a basic system of dimensions and units. After
doing that, this chapter defines properties of a substance and introduces the
concepts of work, power, and energy.
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Demonstrate understanding of thermodynamic properties, and
methods of measuring intensive and extensive properties.
2. Explain the concepts of heat, work, and energy.
TLO 1 Thermodynamic Properties
Overview
Thermodynamic properties describe measurable characteristics of a
substance. These thermodynamic properties often identify substances or
distinguish between different substances.
Thermodynamics is concerned with both thermal and mechanical properties
of substances and their measurement. Such measurements are expressed in
units that characterize the substance under consideration. The operator
must be able to demonstrate understanding of thermodynamic properties
and the methods of measuring intensive and extensive properties.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the following properties:
a. Specific volume
b. Density
c. Specific gravity
d. Humidity
e. Mass
f. Weight
2
Rev 1
g. Intensive
h. Extensive
2. Define the thermodynamic properties of temperature and convert
between the Fahrenheit, Celsius, Kelvin, and Rankine scales.
3. Define the thermodynamic properties of pressure and convert between
all pressure scales.
ELO 1.1 Properties and Definitions
Introduction
The operator is provided with information that is displayed using various
units of measurement. The operator must be able to convert between these
units of measurement to ensure that the plant is operating within established
limits. Instrument readings may provide information in units that are
different from those provided by a procedure. In this case, the operator will
be required to perform a unit conversion. Additionally, documents
provided by equipment vendors may also contain units of measurement that
need conversion from English units to metric units or vice versa.
Measurement Systems
English units and the International System of Units (SI) are the most
commonly used measurement systems available.
English Units of Measurement
The English system, primarily used in the United States, consists of various
units for each of the fundamental dimensions such as length, mass, and
time, shown below in the table.
Length
Mass
Time
Inch
Ounce
Second*
Foot*
Pound*
Minute
Yard
Ton
Hour
Mile
Day
NOTE: *Denotes standard unit of measure
The English, or foot-pound-second (FPS), system is used within the United
States in the engineering field.
Rev 1
3
International System of Units (SI)
The SI system is made up of two related systems, the meter-kilogramsecond (MKS) system, and the centimeter-gram-second (CGS) system.
The MKS and CGS systems each use a decimal-based system in which
prefixes are used to denote powers of ten. For example, one kilometer is
1,000 meters and one centimeter is 0.01 meters. Units of conversion in the
English system are not as straightforward. For example, a mile is 5,280 feet
and a foot is 12 inches.
The MKS system is used primarily for calculations in the field of physics,
while both the MKS and CGS systems are used in the field of chemistry.
The units for each of these systems are shown below in the two tables.
MKS Units of Measurement
Length
Mass
Time
Millimeter
Milligram
Second*
Meter*
Gram
Minute
Kilometer
Kilogram*
Hour
Day
NOTE: *Denotes standard unit of measure
CGS Units of Measurement
Length
Mass
Time
Centimeter*
Milligram
Second*
Meter
Gram*
Minute
Kilometer
Kilogram
Hour
Day
NOTE: *Denotes standard unit of measure
4
Rev 1
Unit Conversions
It is necessary to develop relationships of known equivalents such as
conversion factors to apply measurements between and within the SI and
English systems. These equivalents are used to convert from the given units
of measure to the desired unit of measure.
Conversion Factors
Conversion factors are relationships (or ratios) of equivalent values and are
applied to a given measurement to convert it into the desired units. The
equivalent relationships between different units of measurement are defined
in conversion tables. Some examples are given below in the table.
Typical Conversion Table
Unit
English Units of
Measurement
Meter-Kilogram-Second
(MKS) Units of
Measurement
Length
1 yard (yd)
= 0.9144 meter (m)
12 inches (in.)
= 1 ft
5,280 feet (ft)
= 1 mi
1 (meter) m
= 3.281 ft
1 in.
= 0.0254 m
60 seconds (sec)
= 1 minute (min)
3,600 sec
= 1 hour (hr)
1 pound mass (lbm)
0.4535 kg
2.205 lbm
= 1 kg
1 kilogram (kg)
= 1,000 grams (g)
1 square foot (ft2)
= 144 in.2
10.764 ft2
= 1 square meter (m2)
1 square yard (yd2)
= 9 ft2
1 square mile (mi2)
3.098 x 106 yd2
Time
Mass
Area
Rev 1
5
Unit
English Units of
Measurement
Meter-Kilogram-Second
(MKS) Units of
Measurement
Volume
7.48 gallon (gal)
= 1 cubic foot (ft3)
1 gal
= 3.785 l (liter)
1 liter (l)
= 1,000 cubic centimeters
(cm3)
Performing Unit Conversions
Unit conversion is essentially a multiplication by one, which does not
change the magnitude of the original quantity, but only its measurementunit identity. To convert from one measurement unit to another (for
example, to convert 5 feet to inches), first select the appropriate equivalence
relationship from the conversion table (in this case, 1 ft = 12 in.).
Next, multiply the original quantity by the appropriate conversion factor in
such a manner that the unwanted unit (feet) cancels algebraically and the
desired unit (inches) remains.
1 𝑓𝑡 12 𝑖𝑛𝑐ℎ𝑒𝑠
12 𝑖𝑛𝑐ℎ𝑒𝑠
=
𝑜𝑟 1 =
1 𝑓𝑡
1 𝑓𝑡
1 𝑓𝑡
Thus, 5 feet equals 60 inches.
Steps for Unit Conversion
Step Action
1.
Identify the units given and the units required.
2.
Choose the equivalence relationship(s)* for the two units.
3.
Arrange the equivalence ratio in the appropriate manner:
(
4.
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠
)
𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠
Multiply the quantity by the ratio.
*If you cannot find a conversion relationship between the given
units and the desired units in the conversion tables, you may need to
use multiple conversion factors.
6
Rev 1
Example 1: Convert 795 meters to feet
The following example steps through the process for converting from a
given unit to a desired unit.
Rev 1
7
Solution 1: Convert 795 m to ft.
Step Action
1.
Identify the units given and the units required: meters to feet.
2.
Select the equivalence relationship from the conversion table:
1 𝑚𝑒𝑡𝑒𝑟(𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠) = 3.281 𝑓𝑡(𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠)
3.
Arrange the equivalence ratio in the appropriate manner:
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠
(
)
𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠
1=
4.
3.281 𝑓𝑡
1𝑚
Multiply the quantity by the ratio:
(795 𝑚) (
3.281 𝑓𝑡
795 𝑚 3.281 𝑓𝑡
)=(
)(
)
1𝑚
1
1𝑚
= 795 × 3.281 𝑓𝑡 = 2,608.395 𝑓𝑡
Multiple conversion factors must be used if an equivalence relationship
between the given units and the desired units cannot be found in the
conversion tables. The conversion is performed in several steps, until the
measurement is in the desired units. The given measurement must be
multiplied by each conversion factor; the answer will be in the desired units,
after the common units have been canceled.
Example 2: Convert 2.91 Square Miles to Square Meters
Step Action
1.
Select the equivalence relationship from the conversion table.
Multiple conversions will be necessary because there is no direct
conversion shown for square miles to square meters. For this
example, the following conversions will be used: square miles to
square yards to square feet to square meters.
1 𝑚𝑖 2 = 3.908 × 106 𝑦𝑑 2
1 𝑦𝑑2 = 9 𝑓𝑡 2
10.764 𝑓𝑡 2 = 1 𝑚2
8
Rev 1
Step Action
2.
Express the relationship as a ratio (desired units/present units):
3.098 × 106 𝑠𝑞 𝑦𝑎𝑟𝑑
1=
1 𝑠𝑞 𝑚𝑖𝑙𝑒
3.
Multiply the original quantity by the ratio:
3.098 × 106
(2.91 𝑠𝑞 𝑚𝑖𝑙𝑒𝑠) (
) = 9.015 × 106 𝑠𝑞 𝑦𝑎𝑟𝑑
1 𝑠𝑞 𝑚𝑖𝑙𝑒
4.
Repeat the steps, until the value is in the desired units.
1=
9 𝑠𝑞 𝑓𝑡
1 𝑠𝑞 𝑦𝑎𝑟𝑑
9 𝑠𝑞 𝑓𝑡
(9.015 × 106 𝑠𝑞 𝑦𝑎𝑟𝑑) (
) = 8.114 × 107 𝑠𝑞 𝑓𝑡
1 𝑠𝑞 𝑦𝑎𝑟𝑑
1=
1 𝑠𝑞 𝑚𝑖𝑙𝑒
10.764 𝑠𝑞 𝑓𝑡
1 𝑠𝑞 𝑚𝑖𝑙𝑒
)
10.764 𝑠𝑞 𝑓𝑡
(8.114 × 107 𝑠𝑞 𝑓𝑡) (1 𝑠𝑞 𝑚𝑖𝑙𝑒)
=
10.764 𝑠𝑞 𝑓𝑒𝑒𝑡
(8.114 × 107 𝑠𝑞 𝑓𝑡) (
=
(8.114 × 107 𝑠𝑞 𝑚𝑖𝑙𝑒)
= 7.538 × 106 𝑠𝑞 𝑚𝑖𝑙𝑒𝑠
10.764
All of the conversions can be performed in a single equation, as long as all
of the appropriate conversion factors are included.
(2.91 𝑠𝑞 𝑚𝑖𝑙𝑒𝑠) (
3.098 × 10 𝑠𝑞 𝑦𝑑 9 𝑠𝑞 𝑓𝑡
1 𝑠𝑞 𝑚
)(
)(
)
1 𝑠𝑞 𝑚𝑖𝑙𝑒
1 𝑠𝑞 𝑦𝑑 10.764 𝑠𝑞 𝑓𝑡
=
(2.91)(3.098 × 106 )(9)(1 𝑠𝑞 𝑚)
10.764
=
8.114 × 107 𝑠𝑞 𝑚
0.764
= 7.538 × 107 𝑠𝑞 𝑚
Rev 1
9
Conversion Factors for Common Units of Mass
Unit
Gram
(g)
Kilogram
(kg)
Metric Ton
(t)
Pound-mass (lbm)
1 gram (g)
1
0.001
10-5
2.2046 x 10-3
1 kilogram (kg)
1,000
1
0.001
2.2046
1 metric ton (t)
106
1,000
1
2204.6
1 pound-mass
(lbm)
453.59
0.45359
4.5359 x 10-4
1
1 slug
14.594
14.594
0.014594
32.174
Conversion Factors for Common Units of Length
Unit
Centimete
r (cm)
Meter
(m)
Kilometer
(km)
Inch
(in.)
Foot
(ft)
Mile (mi)
1 centimeter
(cm)
1
0.01
10-5
0.3937
0.03280
8
6.2137 x
10-6
1 meter (m)
100
1
0.001
39.370
3.2808
6.2137 x
10-4
1 kilometer
(km)
105
1,000
1
39,370
3280.8
0.62137
1 inch (in.)
2.5400
0.02540
0
2.5400 x
10-5
1
0.08333
3
1.5783 x
10-5
1 foot (ft)
30.480
0.30480
3.0480 x
10-4
12.000
1
1.8939 x
10-4
1 mile (mi)
1.6093 x
105
1,609.3
1.6093
63,360
5,280
1
Conversion Factors for Common Units of Time
Unit
Second (s)
Minute
(min)
Hour (hr)
Day (d)
Year (yr)
1 second (s)
1
0.017
2.7 x 10-4
1.16 x 10-5
3.1 x 10-8
1 minute
(min)
60
1
0.017
6.9 x 10-4
1.9 x 10-6
10
Rev 1
Unit
Second (s)
Minute
(min)
Hour (hr)
Day (d)
Year (yr)
1 hour (hr)
3,600
60
1
4.16 x 10-2
1.14 x 10-4
1 day (d)
86,400
1,440
24
1
2.74 x 10-3
1 year (yr)
3.15 x 107
5.26 x 105
8,760
365
1
Properties
Thermodynamic properties describe measurable characteristics of a
substance. These properties are used to identify substances or distinguish
between different substances.
Mass and Weight
A body's mass (m) measures the amount of material that is present in the
body. A body's weight is the force exerted by that body, when its mass is
accelerated in a gravitational field. Mass and weight are related by a
variation of Newton's Second Law of Motion, as shown below (in the
English system of units).
𝑚𝑔
𝑤𝑡 =
𝑔𝑐
Where:
wt = weight, in units of pound-force (lbf)
m = mass (lbm)
g = acceleration due to gravity = 32.17 ft/s
𝑔𝑐 = gravitational constant = 32.17 lbm-ft/lbf-s2
This relationship is only true at sea level, where the acceleration due to
gravity is 32.17 ft/sec2.
Note
In the above equation, acceleration (a) is often written as
gravity (g) because the acceleration is the gravitational
acceleration due to the earth’s gravitational field (g =
32.17 ft/sec2).
The weight of a body is the force produced, when the mass of the body is
accelerated by gravity. The mass of a given body remains constant, even if
the gravitational acceleration acting upon that body (and, consequently, its
weight) changes.
Rev 1
11
The English system uses the pound-force (lbf) as the unit of weight. The
basic unit of mass in the English system is the slug; however, the unit of
mass generally used is the pound-mass (lbm), where 1 slug = 32.17 lbm.
The gravitational constant (gc) modifies Newton’s second law, such that 1
lbf of weight is generated by 1 lbm at the surface of the earth. This
relationship is only true at the surface of the earth, however, where the
acceleration due to gravity is exactly 32.17 ft/s2.
Example:
Using Newton’s second law, prove that 1 lbf is equivalent to 1 lbm on the
earth's surface.
Solution:
𝑤𝑡 =
𝑚𝑔
𝑔𝑐
1 𝑙𝑏𝑓 = (1 𝑙𝑏𝑚)(32.17
𝑓𝑡
𝑙𝑏𝑚_ 𝑓𝑡
)/(32.17
)
𝑠2
𝑙𝑏𝑓 _ 𝑠 2
1 𝑙𝑏𝑓 = 1𝑙𝑏𝑓 (for example, an equality)
Specific Volume
The specific volume (v) of a substance is the total volume (V) of that
substance divided by its total mass (m); for example, the volume per unit
mass. It has units of cubic feet per pound-mass (ft3/lbm).
𝑣=
𝑉
𝑚
Where:
v = specific volume (ft3/lbm)
V = volume (ft3)
m = mass (lbm)
Density
The density ρ of a substance is the total mass (m) of that substance divided
by its total volume (V); for example, the mass per unit volume. An object is
said to be very dense if there is a large mass situated within a relatively
small volume. It has units of pound-mass per cubic feet (lbm/ft3). The
density (ρ) of a substance is the reciprocal of its specific volume (v).
12
Rev 1
𝜌=
𝑚 1
=
𝑉 𝑣
Where:
ρ = density (lbm/ft3)
m = mass (lbm)
V = volume (ft3)
v = specific volume (ft3/lbm)
The density of a substance may be varied by changing its pressure and/or
temperature. For example, increasing the pressure on a substance can
increase its density, whereas increasing its temperature will lower its
density.
The effect of pressure on the densities of liquids and solids is relatively
small; however, the effect of temperature on the densities of liquids and
gases can be substantial (as is the effect of pressure on the densities of
gases).
Specific Gravity
Specific gravity (S.G.) identifies the relative density of a substance
compared to the density of water at a standard temperature and pressure
(typically, atmospheric pressure). Physicists use 39.2°F (4°C) as the
standard, since this is the temperature at which water is at its densest state.
The density of water is 1.00 g/cm3 at the standard temperature. The specific
gravity for a liquid has the same numerical value, therefore, as its density
(in units of g/cm3). Specific gravities must be determined and specified at
particular temperatures because the density of a liquid varies with
temperature.
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 (𝑆. 𝐺. ) =
𝜌𝑙𝑖𝑞𝑢𝑖𝑑
𝜌𝑤𝑎𝑡𝑒𝑟
Humidity
Humidity is the amount of moisture, such as water vapor, that is present in a
volume of air. The amount of moisture that can be held in air varies with
the air's temperature, such that warmer air will hold a greater amount of
water vapor.
Humidity can be expressed as an absolute or relative quantity. Absolute
humidity is the mass of water vapor contained within a unit volume of air,
for example, grams of water/cm3 of air. Relative humidity is the actual
amount of water vapor present in air compared to the maximum amount of
vapor that the air could contain at that temperature. Relative humidity is
Rev 1
13
expressed as a percentage; it will be 0 percent if no water vapor is present in
the air and 100 percent if the air is saturated with water vapor (for example,
if the air is holding as much water vapor as possible).
Intensive And Extensive Properties
Thermodynamic properties can be divided into two general classes:
intensive and extensive. Intensive properties are independent of the actual
mass of the substance in question (for example, temperature, pressure,
specific volume, and density), whereas the value of an extensive property
varies directly with the amount mass under consideration (such as mass and
volume).
For example, if a homogeneous quantity of matter in a given state is divided
into two equal parts, each part will have the same values of the intensive
properties as the original, but one-half the values of the extensive
properties.
Knowledge Check
Which thermodynamic property is a measure of relative
density compared to the density of water?
A.
Specific volume
B.
Density
C.
Specific density
D.
Specific gravity
Knowledge Check
Which one of the following is an example of an
extensive thermodynamic property?
14
A.
Temperature
B.
Pressure
C.
Volume
D.
Density
Rev 1
Knowledge Check
A reactor core's thermal power is 2,000 megawatts
thermal (MWth). Convert this to BTU/hr.
A.
6.824E9 BTU/hr
B.
6,824E12 BTU/hr
C.
2.93E9 BTU/hr
D.
2.93E12 BTU/hr
ELO 1.2 Thermodynamic Properties of Temperature
Introduction
Thermodynamics uses several types of temperature scales that operators
must recognize and understand to perform their daily monitoring functions.
Temperature Scales
Temperature is a measure of the molecular activity of a substance (that is,
temperature is a relative measure of how hot or cold a substance is) and can
be used to predict the direction of heat transfer. Higher temperatures result
in greater molecular movement within a substance.
The Fahrenheit (F) and Celsius (C) scales are normally used for temperature
measurement purposes and specify the number of increments between the
freezing and boiling points of water at standard atmospheric pressure. The
Celsius scale has 100 units between these points, and the Fahrenheit scale
has 180 units.
The freezing point of water was selected as the zero point of the Celsius
scale. The freezing point of water on the Fahrenheit scale is 32°F; 0°F
corresponds to the freezing point of a brine-water solution.
The temperature at which water boils was set at 100°C on the Celsius scale
and is 212°F on the Fahrenheit scale; 100°F was historically chosen as the
temperature of a typical human body (approximately 98.6°F). The
relationship between these scales is shown by the following equations and
illustrated below in the figure.
Rev 1
15
Figure: Boiling and Freezing Points of Water for Celsius and Fahrenheit
Temperature Scales
Difference in Scales
The low-temperature condition at which all molecular or atomic motion
ceases is referred to as absolute zero (0) and serves as the basis for two
additional temperature scales. The absolute temperature scale that
corresponds to the Celsius scale is the Kelvin (K) scale and that which
corresponds to the Fahrenheit scale is the Rankine (R) scale. The
relationships between the absolute and relative temperature scales are
shown below in the following figure and equations.

°𝑅 = °𝐹 + 460

°𝐾 = ℃ + 273
Figure: Comparison of Temperature Scales
16
Rev 1
Comparison of Temperature Scales
The conversion from one temperature scale to another is sometimes
required and the operator should be familiar with the process. The
following are examples of temperature scale conversions:
Example 1: Temperature Scale Conversion
What is the Rankine equivalent of 80°C?
Solution 1:
9
℉ = ( ) °𝐶 + 32
5
9
= ( ) (80) + 32
5
= 176°𝐹
°𝑅 = ℉ + 460
= 176 + 460
= 636°𝑅
Example 2: Temperature Scale Conversion
What is the Kelvin equivalent of 80°F?
Solution 2:
5
℃ = ( ) (℉ − 32)
9
5
= ( ) (80 − 32)
9
= 27℃
°𝐾 = ℃ + 273
= 27 + 273
= 300°𝐾
Rev 1
17
Practice Question
The water in the reactor coolant system returning to the
reactor is 550.4°F. What is this temperature in degrees
Celsius, Kelvin, and Rankine?
Knowledge Check
The low-temperature condition at which all molecular or
atomic motion ceases is referred to as
.
A.
reference point
B.
freeze point
C.
absolute zero
D.
reference zero
ELO 1.3 Thermodynamic Properties of Pressure
Introduction
The pressure of a substance is the force exerted by the material (per unit
area) on the boundaries that surround it. This force is caused by the atomic
or molecular collisions that occur between the substance itself and its
boundaries. As the individual particles strike the boundaries, they exert
forces that attempt to push the boundaries outward.
Pressure is typically expressed in units of pounds-force per square inch
(lbf/in.2 or psi) in the English System of Measurement, but may also be
measured or expressed using equivalent columns of liquid, such as water or
mercury. The height of a given column of liquid generates a pressure at its
base that converts to units of force per unit area, per the following equation.
𝑃=
18
𝜌𝑔ℎ
𝑔𝑐
Rev 1
Where:
P = pressure (lbf/in.2 or psi)
ρ = density (lbm/ft3)
g = acceleration due to gravity (32.17 ft/s2)
h = height of column (ft)
𝑔𝑐 = gravitational constant = 32.17 lbm-ft/lbf-s2
This formula requires an additional conversion factor, for completeness (for
example, 1 ft2 = 144 in.2).
Pressure Scales
Pressure is measured relative to a perfect vacuum (for example, the
complete absence of atoms/molecules) is called absolute pressure (psia).
Gauge pressure (psig) is measured relative to atmospheric pressure at sea
level (~ 14.7 psig). Most system-pressure gauges register zero (0), when
open to the atmosphere; hence, pressure gauges actually measure the
pressure difference between the observed substance and the surrounding
atmosphere.
Figure: Pressure Scale Relationships
Pressure below atmospheric pressure is designated as a vacuum. A perfect
vacuum corresponds to an absolute pressure of zero (0) (or 0 psia); all
values of absolute pressure, therefore, are positive. Gauge pressures are
positive when above atmospheric pressure, and negative when below. The
figure below shows the relationships between absolute, gauge, vacuum, and
atmospheric pressures.
Example 1 Practice:
Rev 1
19
A pressure gauge on a condenser reads 27 inches of mercury (Hg) vacuum.
What is the absolute pressure corresponding to this vacuum (assume an
atmospheric pressure of 15 psia)?
A.
14.0 psia
B.
13.5 psia
C.
1.5 psia
D.
1.0 psia
Figure: Gauge and Absolute Pressure Scale Relationships
Solution – Practice Example 1:
𝑥 𝑝𝑠𝑖𝑣
27 𝑖𝑛𝑐ℎ𝑒𝑠 𝐻𝑔
=
15 𝑝𝑠𝑖𝑣
29.9 𝑖𝑛𝑐ℎ𝑒𝑠 ℎ𝑔
=
27 𝑖𝑛 𝐻𝐺 ×15 𝑝𝑠𝑖𝑣
29.9 𝑖𝑛 𝐻𝑔
= 13.5 psiv
13.5 psiv = 1.5 psia
Example 2 Practice:
Which one of the following is arranged from the lowest pressure to the
highest pressure?
20
A.
2 psig, 12 inches Hg absolute, 8 psia
B.
2 psig, 18 inches Hg absolute, 8 psia
C.
12 psia, 20 inches Hg absolute, 2 psig
D.
12 psia, 30 inches Hg absolute, 2 psig
Rev 1
Figure: Comparison of Pressure Ranges
As discussed previously, pressure also can be measured in terms of an
equivalent column of liquid, usually water, or mercury, which is referred to
as hydrostatic pressure. Hydrostatic pressure at a given reference point is
the product of a fluid’s density, its height above a reference point, and the
acceleration due to gravity.
The greater the density or height of the fluid column, the more pressure it
exerts on a given area. The most common units of this type of measurement
are feet of water (ft H2O), inches of mercury (in. Hg), and millimeters of
mercury (mm Hg). The different hydrostatic pressure measurements can be
compared using the following relationships, such as conversion factors:


0.491 psi = 1 inch Hg
0.433 psi = 1 ft H2O
Example 3: Pressure Relationships
How deep can a diver descend in ocean water (density = 64 lbm/ft3) without
damaging his watch, which will withstand an absolute pressure of 80 psia?
Solution 3:
Assume Patm = 14.7 psia
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑢𝑔𝑒
80 𝑝𝑠𝑖𝑎 = 14.7 𝑝𝑠𝑖𝑎 + 𝑃𝑔𝑎𝑢𝑔𝑒
𝑃𝑔𝑎𝑢𝑔𝑒 = (80 𝑝𝑠𝑖𝑎 − 14.7 𝑝𝑠𝑖𝑎) = 65.3 𝑝𝑠𝑖𝑔
𝑃𝑔𝑎𝑢𝑔𝑒 =
65.3
𝜌𝑔ℎ
𝑔𝑐
𝑙𝑏𝑓
=
𝑖𝑛2
Rev 1
(64
𝑓𝑡
𝑙𝑏𝑚
) (32.17 2 ) (ℎ)
3
𝑓𝑡
𝑠
_
𝑙𝑏𝑚 𝑓𝑡
(32.17
)
𝑙𝑏𝑚_ 𝑠 2
21
𝑙𝑏𝑓
𝑓𝑡
) (32.17 2 )
𝑖𝑛2
𝑖𝑛2
𝑠
ℎ=[
] (144 2 )
𝑓𝑡
𝑙𝑏𝑚
𝑓𝑡
(64 3 ) (32.17 2 )
𝑓𝑡
𝑠
(65.3
ℎ = 1.47 × 102 𝑓𝑡
Pressure Relationships
Patm is atmospheric pressure, which is also called the barometric pressure.
Pgauge indicates the gauge pressure, and Pvac is vacuum. The following
examples relating the various pressures will be helpful in understanding the
idea of gauge versus absolute pressures:
Example 3: Pressure Relationships
How deep can a diver descend in ocean water (density = 64 lbm/ft3) without
damaging his watch, which will withstand an absolute pressure of 80 psia?
(P = density x height)
Solution 3: Assume
𝑃𝑎𝑡𝑚 = 14.7 𝑝𝑠𝑖𝑎
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑢𝑔𝑒
80 𝑝𝑠𝑖𝑎 = 14.7 + 𝑃𝑔𝑎𝑢𝑔𝑒
𝑃𝑔𝑎𝑢𝑔𝑒 = (80 − 14.7) = 65.3 𝑝𝑠𝑖𝑔
𝑃𝑔𝑎𝑢𝑔𝑒 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × ℎ𝑒𝑖𝑔ℎ𝑡 = 𝜌𝐻
(65.31
𝑙𝑏𝑓
𝑖𝑛2
𝑙𝑏𝑚
)
(144
) = (64 3 ) 𝐻
2
2
𝑖𝑛
𝑓𝑡
𝑓𝑡
(65.3)(144)
64
𝐻 = 146.9 𝑓𝑡
𝐻=
Example 4: Pressure Relationships
What is the absolute pressure at the bottom of a swimming pool 6 feet deep
that is filled with fresh water?
Assume 𝑃𝑎𝑡𝑚 = 14.7 𝑝𝑠𝑖𝑎
Solution 4:
𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑢𝑔𝑒
= 14.7 + 𝜌𝐻 (𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × ℎ𝑒𝑖𝑔ℎ𝑡)
22
Rev 1
𝑙𝑏𝑚
(62.4 3 ) (6 𝑓𝑡. )
𝑓𝑡
= 14.7 + [
]
𝑖𝑛2
144 2
𝑓𝑡
= 14.7 + 2.6
𝑃𝑎𝑏𝑠 = 17.3 𝑝𝑠𝑖𝑎
In addition to pounds per square inch, pressure can be measured with
reference to the force that exists in a column of fluid at a certain height.
The most common of these are inches of water, inches of mercury, and
millimeters of mercury. Conversion factors are listed below:

14.7 psia = 408 inches of water

14.7 psia = 29.9 inches of mercury

1 inch of mercury = 25.4 millimeters of mercury
Knowledge Check
Fill in the blanks below:
1. 40 inches Hg (absolute) = _____ psia or ______ psig
2. 20 ft of water (gauge) = _______ psig or _______
psia
3. 13 psiv = _______ psia or ______ inches of Hg
(vacuum)
4. 28 inches of Hg (vacuum) = _______ psia or _______
psiv
5. 5 inches of water (gauge) = _______ ft of water
(gauge) or _______ psig
Rev 1
A.
(1) 19.6, 4.9; (2) 8.7, 23.4;(3) 1.7, 26.4; (4) 1.0,
13.7;(5 ) 0.42, 0.18
B.
(1) 19.4, 4.9; (2) 8.7, 22.4;(3) 1.6, 24.4; (4) 1.0,
13.7;(5) 0.42, 0.18
C.
(1) 20.6, 4.9; (2) 9.7, 23.4;(3) 1.6, 26.4; (4) 1.0,
13.7;(5) 0.42, 0.18
D.
(1) 19.6, 4.9; (2) 8.7, 22.3;(3) 1.7, 26.4; (4) 1.0,
13.7;(5) 0.42, 0.18
23
TLO 1 Summary
1. Define the following properties








Mass (m) is the measure of the amount of material present in that
body.
Weight (wt) is the force exerted by that body when its mass is
accelerated in a gravitational field.
Specific volume (v) is the total volume (V) of a substance divided
by the total mass (m) of that substance.
Density (ρ) is the total mass (m) of a substance divided by the total
volume (V) occupied by that substance.
Specific gravity (S.G.) is a measure of the relative density of a
substance compared to the density of water at a standard
temperature.
Humidity is the amount of moisture, such as water vapor, in air. It
can be measured in absolute or relative units.
Intensive properties are properties that are independent of the
amount of mass.
Extensive properties are those that vary directly with mass.
2. Define the thermodynamic properties of temperature and convert
between the Fahrenheit, Celsius, Kelvin, and Rankine scales.


Temperature is a measure of the molecular activity of a substance.
o Absolute zero = -460°F or -273°C
o Freezing point of water = 32°F or 0°C
o Boiling point of water = 212°F or 100°C
Conversions between the different scales can be made using the
following formulas:
o °F = 32 + (9/5)°C
o °C = (°F - 32)(5/9)
o °R = °F + 460
o °K = °C + 273
3. Define the thermodynamic properties of pressure and convert between
all pressure scales.

Pressure is a measure of the force per unit area exerted on the
boundaries of a substance (or system).
o 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝑃𝑔𝑎𝑢𝑔𝑒
o 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 − 𝑃𝑣𝑎𝑐

24
Converting between the different pressure units can be done using
the following conversions:
o 14.7 𝑝𝑠𝑖𝑎 = 34 𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 408 𝑖𝑛. 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
o 14.7 𝑝𝑠𝑖𝑎 = 29.9 𝑖𝑛. 𝑜𝑓 𝑚𝑒𝑟𝑐𝑢𝑟𝑦
o 1 𝑖𝑛. 𝑜𝑓 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 = 0.0254 𝑚 𝑜𝑓 𝑚𝑒𝑟𝑐𝑢𝑟𝑦
Rev 1
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Define the following properties:
a. Specific volume
b. Density
c. Specific gravity
d. Humidity
e. Mass
f. Weight
g. Intensive
h. Extensive
2. Define the thermodynamic properties of temperature and convert
between the Fahrenheit, Celsius, Kelvin, and Rankine scales.
3. Define the thermodynamic properties of pressure and convert between
all pressure scales.
TLO 2 Concepts of Heat, Work, and Energy
Overview
Thermodynamics is the branch of science that deals with energy and the
transformation of energy from one form to another. It is necessary to
understand some basic energy concepts and terminology to gain a complete
understanding of many thermodynamic topics.
This lesson will identify and explain the different forms of energy, examine
the conversion of energy from one form to another, and explain concepts
and terminology related to energy, work, and power. In addition, the lesson
will describe the relationship between energy, work, and power, including
the equation that allows us to evaluate those relationships.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the following thermodynamic properties:
a. Potential energy
b. Kinetic energy
c. Specific internal energy
d. Specific p-v energy
e. Specific enthalpy
f. Entropy
2. Explain the relationship between work, energy, and power and define
the following:
a. Heat
b. Latent heat
c. Sensible heat
d. Unit used to measure heat
Rev 1
25
e. Specific heat
f. Super heat
3. Explore the transfer of heat between substances as a part of
thermodynamic processes and cycles.
ELO 2.1 Thermodynamic Properties of Energy
Introduction
Heat and work are two ways in which energy can be transferred across the
boundaries of a system.
Energy
Energy is defined as the capacity to produce an effect, for example, perform
work or produce heat. It is difficult to define energy in a general sense, but
easier to define it in terms of the work done on or by a system. Specifically,
a given system possesses a certain quantity of energy that is decreased when
the system does work (on its surroundings or on another system) and
increased when work is done on the system.
A working fluid is a substance that receives, transfers, and transmits energy
in a thermodynamic system. In most systems, the working substance is a
fluid (such as a liquid or gas). For example, water is the working fluid in a
nuclear steam supply system. Fluid energy is decreased when work is done
on the fluid and increased when the fluid does work.
There are many different forms of energy, such as mechanical energy,
thermal energy, electrical energy, chemical energy, and nuclear energy.
The total energy of a substance, which is always conserved, is the sum of
the various forms of energy that the substance possesses.
Thermo-mechanical energy is classified as either stored energy (such as
energy contained within the mass) or transient energy (energy associated
with the conversion from one stored form to another or the transition from
one system to another).
The four forms of stored energy that are possessed by the working fluid in a
typical energy-transfer system are potential energy, kinetic energy, internal
energy, and PV (flow) energy. The two forms of transient energy are work
and heat.
Potential Energy
Potential energy (PE) is the energy that a substance possesses as a function
of its position relative to a given reference point. The amount of PE an
object contains is dependent upon its mass (m) and elevation relative to the
reference location (z). Potential energy will exist whenever an object that
has mass is positioned within a force field. The most common example is
an object situated within the earth's gravitational field, as shown below:
26
Rev 1
Figure: An Object Acted On By The Earth’s Gravity
Examples of Potential Energy
Using English system units, PE is defined as follows:
𝑃𝐸 =
𝑚𝑔𝑧
𝑔𝑐
Where:
PE = potential energy (ft-lbf)
m = mass (lbm)
z = height above some reference level (ft)
g = acceleration due to gravity (ft/s2)
gc = gravitational constant = 32.17 ft-lbm/lbf-s2
The acceleration due to gravity (g) is numerically equal to the gravitational
constant (gc), although the units differ. Thus, the potential energy (in. ftlbf) numerically equals the product of the mass (in. lbm) and the height (in.
ft) above some reference level.
Example:
Determine the potential energy of 50.0 lbm of water that is located in a
storage tank 100 ft above the ground.
Rev 1
27
Solution:
𝑃𝐸 =
𝑚𝑔𝑧
𝑔𝑐
𝑓𝑡
) (100 𝑓𝑡)
𝑠2
𝑓𝑡 _ 𝑙𝑏𝑚
32.17
𝑙𝑏𝑓 _ 𝑠 2
(50.0 𝑙𝑏𝑚) (32.17
𝑃𝐸 =
𝑃𝐸 = 5.00 × 103 𝑓𝑡 _ 𝑙𝑏𝑓
Occasionally, in thermodynamics, energy is expressed in units of BTU
(British thermal units). To convert to BTU, the following conversion factor
is applied.
778 𝑓𝑡 _ 𝑙𝑏𝑓 = 1 𝐵𝑇𝑈, or 1 = (778
𝑓𝑡 _ 𝑙𝑏𝑓
𝐵𝑇𝑈
)
Example:
Determine the potential energy (in BTU) associated with 1.0 lbm of water at
an elevation of 50 feet above a reference height.
Solution:
𝑃𝐸 =
𝑚𝑔𝑧
𝑔𝑐
𝑓𝑡
(1.0 𝑙𝑏𝑚) (32.17 2 ) (50 𝑓𝑡)
𝑠
𝑃𝐸 = [
]
_
𝑓𝑡 𝑙𝑏𝑚
32.17
𝑙𝑏𝑓 _ 𝑠 2
To convert to BTUs, divide by (BTU/778 ft-lbf), so
𝑃𝐸 = 0.064 𝐵𝑇𝑈 or 6.4 × 10−2 𝐵𝑇𝑈
Kinetic Energy
Kinetic energy (KE) is the energy that a body possesses as a result of its
relative motion and may be defined as the energy needed to accelerate a
body from rest to its current velocity. The body will maintain this kinetic
energy, unless it experiences a new net force.
Example of Kinetic Energy
𝐾𝐸 =
𝑚𝑣 2
2𝑔𝑐
Where:
28
Rev 1
KE = kinetic energy (ft-lbf)
m = mass (lbm)
v = velocity (ft/s)
gc = gravitational constant = 32.17 ft-lbm/lbf-s2
Example:
Determine the kinetic energy of 7 lbm of steam flowing through a pipe at a
velocity of 100.0 ft/sec.
Solution: Using Equation
𝑚𝑣 2
𝐾𝐸 =
2𝑔𝑐
𝑓𝑡 2
𝑠)
𝐾𝐸 =
𝑓𝑡 _ 𝑙𝑏𝑚
(2) (32.17
)
𝑙𝑏𝑓 _ 𝑠 2
(7 𝑙𝑏𝑚) (100.0
𝐾𝐸 = 1.088 × 103 𝑓𝑡 _ 𝑙𝑏𝑓
Or, in terms of BTU:
𝐵𝑇𝑈
𝐾𝐸 = (1.088 × 103 𝑓𝑡 _ 𝑙𝑏𝑓) (
)
778 𝑓𝑡 _ 𝑙𝑏𝑓
𝐾𝐸 = 1.398 𝐵𝑇𝑈
Internal Energy
Both potential and kinetic energies exist as macroscopic (for example,
large-scale) forms of energy that can be observed in terms of the positions
and velocities of objects. A substance possesses several microscopic forms
of energy in addition to these macroscopic forms, which include those due
to the rotational, vibrational, and translational energies of the individual
atoms or molecules in a substance.
These microscopic forms of energy are not easily measured or evaluated
directly; hence, the change in their combined total is generally evaluated
instead. These microscopic forms of energy are collectively referred to as
internal energy, which is customarily represented by the symbol U and
expressed in units of BTU.
The specific internal energy (u) of a substance is its internal energy per unit
mass, which is an intensive property (that is, independent of mass). This
property is equal to the total internal energy (U) divided by total mass (m).
Rev 1
29
𝑢=
𝑈
𝑀
Where:
u = specific internal energy (BTU/lbm)
U = internal energy (total BTU)
m = mass (lbm)
Example:
Determine the specific internal energy of 12 lbm of steam, if its total
internal energy is 2.300 x 104 BTU.
Solution: Using Equation for Specific Internal Energy
𝑢=
𝑈
𝑀
2.300 × 104 𝐵𝑇𝑈
𝑢=
12 𝑙𝑏𝑚
103 𝐵𝑇𝑈
𝑢 = 1.917 ×
𝑙𝑏𝑚
PV (Flow) Energy
PV energy (also referred to as flow energy) is important to the
understanding of energy-transfer systems and is numerically equal to the
product of a system's pressure and volume. When the volume of an
enclosed substance is permitted to expand, work is performed on its
surroundings; hence, a fluid under pressure has the capacity to perform
work. The units of PV energy are (lbf/ft2)(ft3), which is equivalent to ft-lbf
(as with other forms of energy, such as PE).
The specific PV energy of a substance is its PV energy per unit mass and
equals the product of the system's pressure and volume divided by the total
mass (or the product of the pressure and specific volume).
𝑃𝑣 =
𝑃𝑉
𝑚
Where:
P = pressure (lbf/ft2)
V = volume (ft3)
v = specific volume (ft3/lbm) = V/m
30
Rev 1
m = mass (lbm)
Example:
Determine the specific PV energy of 15.0 lbm of steam at 1,000 psia in an
18-ft3 tank.
Solution: Using Equation for Flow Energy
𝑃𝑣 =
𝑃𝑉
𝑚
𝑙𝑏𝑓
(1,000 2 ) (18 𝑓𝑡 3 )
𝑖𝑛2
𝑖𝑛
𝑃𝑣 = [
] (144 2 )
15.0 𝑙𝑏𝑚
𝑓𝑡
𝑃𝑣 = 1.73 × 105
𝑓𝑡 _ 𝑙𝑏𝑓
𝑙𝑏𝑚
Enthalpy
Enthalpy (symbolized by H) is a thermodynamic property of a system that
is equivalent to the sum of its internal and flow energies, for example, H =
U + PV.
Specific enthalpy (h) is defined as the total enthalpy per unit mass;
𝐻
therefore, ℎ = 𝑚 = 𝑢 + 𝑃𝑣, where u is the specific internal energy (in.
BTU/lbm), P is the pressure (in. lbf/ft2), and v is the specific volume (in.
ft3/lbm).
Unlike pressure, temperature, and volume, enthalpy is a property of a
substance that cannot be measured directly; rather, the enthalpy of a
substance is measured with respect to a reference value. For example, the
specific enthalpy of water is based on a reference value of zero (0) at
0.01°C and normal atmospheric pressure (~14.7 psia). It is the change in
specific enthalpy, rather than the absolute value, that is of importance in
practical problems.
Entropy
Entropy (symbolized by S) is a measure of a system's inefficiency for doing
work, for a given amount of heat transferred. Specific entropy (s = S/m) is
useful in determining the amount of heat transferred to or from a system
that can and cannot be used to perform work. The change in entropy is
represented by ΔS (or Δs), as in the following relationships.
∆𝑆 =
Rev 1
∆𝑄
𝑇𝑎𝑏𝑠
31
∆𝑠 =
∆𝑞
𝑇𝑎𝑏𝑠
Where:
∆𝑆 = the change in entropy of a system during some process (BTU/°R)
∆𝑄 = the amount of heat transferred to or from the system during the
process (BTU)
Tabs = the absolute temperature at which the heat was transferred (°R)
∆𝑠 = the change in specific entropy of a system during some process
(BTU/lbm-°R)
∆𝑞 = the amount of heat transferred to/from the system during the process
(BTU/lbm)
Like enthalpy, entropy cannot be measured directly, but rather is measured
with respect to a reference value. For example, specific entropy of water is
considered to be zero (0) at 32.02°F and 14.7 psia, which define the
reference condition. While the absolute value of specific entropy may be
unknown, it is the change in specific entropy (Δs) that is important when
solving practical problems.
Knowledge Check
___________ is the measure of energy content of the
fluid due to its temperature, pressure, and volume.
A.
Entropy
B.
Kinetic energy
C.
Enthalpy
D.
Specific internal energy
ELO 2.2 Relationship Between Work, Energy, and Power
Introduction
The purpose of a nuclear-powered generating station is to transfer the
thermal energy produced in the nuclear fuel to the turbine-generator where
the thermal energy is converted into mechanical work and then electrical
energy. Heat is a form of energy in transition and is caused by a difference
in temperature. Work is defined as the force used to move a mass,
multiplied by the distance that the mass was moved. Power is defined as
32
Rev 1
the rate of doing work (that is, the work done per unit time).Each of these
terms is related and must be understood to solve thermodynamic problems.
This lesson will identify and explain the different forms of work, examine
the conversion of energy from one form to another, and explain concepts
and terminology related to work and power. In addition, the lesson will
describe the relationship between energy, work, and power, including the
equation that allows us to evaluate those relationships.
Work
Kinetic, potential, internal, and PV energies are stored forms of energy and
considered properties of a system. Work is also a form of energy, but is
considered a form of energy in transit (like heat) and, consequently, is not a
system property; hence, work is a process done by or on a system, although
a system contains no work. It is important to understand the distinction
between the stored forms of energy, which are properties of a system, and
those forms of energy that are transferred to and from a system.
In mechanical systems, work is defined as the action of a force on an object
through a distance and equals the product of the force (F) and its
displacement (d).
𝑊 = 𝐹𝑑
Where:
W = work (ft-lbf)
F = force (lbf)
d = displacement (ft)
Example:
Determine the amount of work done, if 150 lbf is applied to an object until
it has moved a distance of 30.0 feet.
Solution Using Work Equation:
𝑊 = 𝐹𝑑
𝑊 = (150 𝑙𝑏𝑓)(30.0 𝑓𝑡)
𝑊 = 4.50 × 103 𝑓𝑡 _ 𝑙𝑏𝑓
It is important to distinguish between the work done by a system and that
done on a system by its surroundings. For example, work is done by a
system when its steam is used to turn a turbine generate electricity in a
turbine-generator or a pump is used to move the working fluid from one
location to another. Note that when work is done by a system, the process
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33
leads to a lowering of the system's total stored energy; hence, work done by
a system is considered to be a negative quantity.
Conversely, when work is done by a system on its surroundings, it is
considered to be a positive quantity since the system's total energy increases
as a result. In either case, energy is being converted from one stored form
to another as work is done. The actual amount of work done on or by a
system depends upon the specific process and not simply on the initial and
final conditions of the system (as is the case for the stored forms of energy).
Consider the below figure of a piston/cylinder arrangement. A frictionless
piston is fitted within a vertical cylinder that contains compressed air. If the
piston is released, it will rise in the cylinder until a pressure equilibrium is
attained. Since a force has acted through a distance, work has been done by
this process. Some of the energy that originally existed as stored forms
energy in the pressurized air was converted into potential energy of the
piston's mass. The mechanical work accomplished by the system (such as,
a negative value) is a manifestation of a transfer of stored energy.
Figure: Work Done By a Piston
In the English system of units, the foot-pound force (ft-lbf) is used for work
processes; the SI units are the Newton-meter (Nm) or joule (J).
In thermodynamics, we are primarily concerned with two types of work:
mechanical work and flow (or PV) work.
Flow Work
Flow work (or flow energy) is the work that is required to maintain a
continuous, steady flow of fluid. Flow work is important, when it becomes
necessary to move a fluid from one point to another.
34
Rev 1
Figure: Pipe Boundary Volume For Flow Energy
When a volume of fluid is forced past a boundary in a pipe, as shown in the
figure above, the fluid performs work. This flow work is equivalent to a
force acting through a distance (such as length).
𝑊𝑓𝑙𝑜𝑤 = 𝐹𝐿 (𝑓𝑜𝑟𝑐𝑒 × 𝑙𝑒𝑛𝑔𝑡ℎ)
Since 𝐹𝑜𝑟𝑐𝑒 = 𝑃𝐴 (𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 × 𝑎𝑟𝑒𝑎), 𝑊𝑓𝑙𝑜𝑤 = 𝑃𝐴𝐿
Since 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐴𝐿 (𝑎𝑟𝑒𝑎 × 𝑙𝑒𝑛𝑔𝑡ℎ), 𝑊𝑓𝑙𝑜𝑤 = 𝑃𝑉
Where:
Wflow = flow work (ft-lbf)
P = pressure (lbf/ft2)
V = volume (ft3)
F = force (lbf)
A = area (ft2)
L = length (ft)
Flow work is a form of mechanical work and can also be expressed in units
of BTU. Flow work is also called flow energy. The flow energy per unit
mass of material (for example, w = W/m) is equivalent to Pv.
Energy and Power Equivalences
Three types of units are normally used to measure energy:
1. Mechanical units, such as the foot-pound-force (ft-lbf);
2. Thermal units, such as the British Thermal Unit (BTU); and
3. Electrical units, such as the watt-second (W-sec).
In the MKS and CGS systems, the mechanical units of energy are the joule
(J) and the erg, the thermal units are the kilocalorie (kcal) and the calorie
(cal), and the electrical units are the watt-second (W-sec) and the erg.
Although the units of the various forms of energy are different, they are
equivalent.
Rev 1
35
J. P. Joule, an English physicist, showed that one kilocalorie equals 4,186
joules. These same experiments, when performed using English system
units, show that one British Thermal Unit (BTU) equals 778.3 ft-lbf. These
experiments established the equivalence of mechanical and thermal energy.
Other experiments established the equivalence of electrical energy with
both mechanical and thermal energy. These equivalences are expressed by
the following relationships for engineering applications.



1 𝑓𝑡 _ 𝑙𝑏𝑓 = 1.286 × 10−3 𝐵𝑇𝑈 = 3.766 × 10−7 𝑘𝑊 _ ℎ𝑟
1 𝐵𝑇𝑈 = 778.3 𝑓𝑡 _ 𝑙𝑏𝑓 = 2.928 × 10−4 𝑘𝑊 _ ℎ𝑟
1 𝑘𝑊 _ ℎ𝑟 = 3.413 × 103 𝐵𝑇𝑈 = 2.655 × 106 𝑓𝑡 _ 𝑙𝑏𝑓
The horsepower-hour (hp-hr) is another unit of energy encountered in
engineering applications. It is a mechanical unit of energy defined by the
following relationship:

1 ℎ𝑝 _ ℎ𝑟 = 1.980 × 106 𝑓𝑡 _ 𝑙𝑏𝑓
These relationships can be used to convert various forms of energy between
the English system units.
Most computations involving the working fluid energy in an energy transfer
system are performed in BTUs. Forms of mechanical energy such as
potential energy, kinetic energy, and mechanical work and other forms of
energy such as P-V energy, are usually given in foot-pounds-force. These
are converted to BTUs by using 1 𝐵𝑇𝑈 = 778.3 𝑓𝑡 _ 𝑙𝑏𝑓.
Because this conversion factor is used frequently, a constant called the
mechanical equivalent of heat, usually denoted by the symbol J and
sometimes referred to as Joules constant, is defined as:
𝐽 = 778
𝑓𝑡 _ 𝑙𝑏𝑓
𝐵𝑇𝑈
Example 1:
A pump provides a flow rate of 10,000 liters per minute (lpm). The pump
performs 1.5 x 108 joules of work every 100 minutes. What is the power of
the pump?
Solution 1:
𝑃𝑜𝑤𝑒𝑟 =
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
1.5 × 108 𝑗 1 𝑚𝑖𝑛
𝑃𝑜𝑤𝑒𝑟 = (
)(
)
100 𝑚𝑖𝑛
60 𝑠𝑒𝑐
𝑃𝑜𝑤𝑒𝑟 = 25,000 𝑊𝑎𝑡𝑡𝑠
36
Rev 1
Example 2:
A boy rolls a ball with a steady force of 1 lbf, giving the ball a constant
velocity of 5 ft/s.
What is the power used by the boy in rolling the ball?
Solution 2:
𝑃=
𝐹𝑣
500
(1 𝑙𝑏𝑓) (5
𝑃=
𝑓𝑡
𝑠𝑒𝑐 )
550
𝑃 = 9.0 × 10−3 ℎ𝑝
Example 3:
A racecar traveling at constant velocity can go one-quarter mile (1,455 ft) in
5 seconds. If the motor is generating a force of 1,890 lbf pushing the car,
what is the power of the motor in hp? Assume the car is already at full
speed at t = 0.
Solution 3:
𝑃=
𝐹𝑑
𝑡
(1,890 𝑙𝑏𝑓)(1455 𝑓𝑡)
1 ℎ𝑝
𝑃=(
)(
)
𝑓𝑡 _ 𝑙𝑏𝑓
5 𝑠𝑒𝑐
550 𝑠𝑒𝑐
𝑃 = 1,000 ℎ𝑝
𝑓𝑡
(1,890 𝑙𝑏𝑓) (291
𝑠𝑒𝑐 )
𝑃=(
)
550
𝑃 = 1,000 ℎ𝑝
Power
Power is defined as the time rate of doing work and is equivalent to the rate
of energy transfer. Power has units of energy per unit time and may be
expressed in various units, which have established equivalences. In the
English system, the mechanical units of power are ft-lbf/s or ft-lbf/hr and
hp, the thermal units are BTU/hr, and the electrical units are watts (W) or
Rev 1
37
kilowatts (kW = 103 W). For engineering applications, the equivalence of
these units is expressed by the following relationships:
𝑓𝑡 _ 𝑙𝑏𝑓
𝐵𝑇𝑈

1

1

1 𝑘𝑊 = 3.413 × 103
𝑠
𝐵𝑇𝑈
ℎ𝑟
= 4.6263
= 0.2162
ℎ𝑟
𝑓𝑡 _ 𝑙𝑏𝑓
𝑠
= 1.356 × 10−3 𝑘𝑊
= 2.931 × 10−4 𝑘𝑊
𝐵𝑇𝑈
ℎ𝑟
= 737.6
𝑓𝑡 _ 𝑙𝑏𝑓
𝑠
Horsepower is related to ft-lbf/sec by the following relationship:

1 ℎ𝑝 = 550.0
𝑓𝑡 _ 𝑙𝑏𝑓
𝑠
In SI units, power is measured in watts or joules (W or J). These
relationships can be used to convert between the English and SI systems.
Knowledge Check
A 600 lbm casting is lifted 4 feet to the bed of a milling
machine. How much work is done?
A.
240 ft-lbs
B.
150 ft-lbs
C.
2,400 ft-lbs
D.
1,500 ft-lbs
ELO 2.3 Thermodynamic Properties: Introduction to Heat
Introduction
Many thermodynamic analyses involve the transfer of heat between systems
and/or substances, via thermodynamic processes and cycles, which will be
discussed in later lessons. Heat, as previously stated, is energy in transition
that is caused by a difference in temperature.
Heat
Heat (Q), like work, is energy in transit; however, this form of energy
transfer occurs at the molecular level as a result of a temperature difference.
The BTU is the preferred unit of heat, in most steam-plant applications, and
it is the amount of energy that is necessary to raise the temperature of 1 lbm
of water by 1°F (conventionally, from 59.5°F to 60.5°F).
Like work, the actual amount of heat transferred to or from a system
depends upon the specific process and not simply on the initial and final
conditions of the system (as is the case for stored forms of energy). When
analyzing heat-transfer processes, a positive value for heat indicates that
38
Rev 1
heat is being added to the system (and, consequently, the system's total
energy rises); a loss of heat is denoted by a negative heat value.
The symbol q is used to indicate the amount of heat added to or removed
from a system per unit mass. It is equal to the total heat (Q) added or
removed, divided by the mass (m). The term specific heat is not used for q,
because this term has historically been used for another parameter (see
below).
𝑞=
𝑄
𝑚
Where:
q = heat transferred per unit mass (BTU/lbm)
Q = heat transferred (BTU)
m = mass (lbm)
Example:
Determine the heat transferred per unit mass if 1,500 BTU are transferred to
40.0 lbm of water.
Solution Using Heat Equation
𝑞=
𝑄
𝑚
𝑞 =
1,500 𝐵𝑇𝑈
40.0 𝑙𝑏𝑚
𝑞 = 37.5
𝐵𝑇𝑈
𝑙𝑏𝑚
Sensible Heat
A common means of quantifying the amount of heat that is added to or
removed from a system is to observe the change in the system's temperature
because of the heat-transfer process.
The temperature of a substance always increases when it is heated and
decreases when heat is removed (assuming no phase change accompanies
the process). The heat added to or removed from a substance that produces
a change in its temperature is referred to as sensible heat.
Rev 1
39
Latent Heat
Latent heat is the heat added to or removed from a substance resulting in a
change of phase, but no change in temperature. There are three specific
forms of latent heat, one for each type of phase change.
The latent heat of fusion/freezing is associated with the heat added to or
removed from a substance to change it from a solid to a liquid (and vice
versa). The latent heat of vaporization/condensation is associated with the
liquid-vapor phase change; and the latent heat of sublimation and/or
desublimation is associated with the solid-vapor phase change.
Specific Heat
The magnitude of the temperature change for a given amount of heat
transferred differs from one substance to another. The ratio of the heat (Q)
added to or removed from a substance to the resulting change in
temperature (∆T) is referred to as the substance's heat capacity (Cp); the
specific heat (or specific heat capacity) of a substance (cp) is the heat
capacity per unit mass (that is, cp = Cp /m). The subscript p indicates that
the heat-transfer processes in question occurred under constant-pressure
conditions.
𝐶𝑝 =
𝑄
∆𝑇
𝐶𝑝 =
𝑄
𝑚∆𝑇
𝐶𝑝 =
𝑞
∆𝑇
Where:
Cp = heat capacity at constant pressure (BTU/°F)
cp = specific heat at constant pressure (BTU/lbm-°F)
Q = heat transferred (BTU)
q = heat transferred per unit mass (BTU/lbm)
m = mass (lbm)
∆𝑇 = temperature change (°F)
Example:
How much heat is required to raise the temperature of 5.00 lbm of water
from 50°F to 150°F? Assume that the specific heat (cp) for water is 1.0
BTU/lbm-°F.
40
Rev 1
Solution:
𝐶𝑝 =
𝑄
𝑚∆𝑇
𝑄 = 𝑚𝐶𝑝 ∆𝑇
1.0 𝐵𝑇𝑈
𝑄 = (5 𝑙𝑏𝑚) (
) (150℉ − 50℉)
𝑙𝑏𝑚_ ℉
1.0 𝐵𝑇𝑈
𝑄 = (5 𝑙𝑏𝑚) (
) (100℉)
𝑙𝑏𝑚_ ℉
𝑄 = 5.0 × 102 𝐵𝑇𝑈
Super heat
The number of degrees a vapor is above the saturation temperature (boiling
point) at a specific pressure.
Knowledge Check
Which of the following must be added to or removed
from a substance to produce a temperature change?
Rev 1
A.
Latent heat
B.
Specific heat
C.
Sensible heat
D.
Thermal heat
41
TLO 2 Summary
1. Define the following thermodynamic properties:
 Energy is the capacity of a system to perform work or produce
heat.
 Potential energy (PE) is the energy of position.
 Kinetic energy (KE) is the energy that a body possesses due to its
motion.
 Latent heat is the amount of heat added or removed to produce
only a phase change.
 Sensible heat is heat added or removed that causes a temperature
change.
 Specific enthalpy (h) — ℎ = 𝑢 + 𝑃𝜈, where u is the specific
internal energy (BTU/lbm) of the system being studied, P is the
pressure of the system (lbf/ft2), and ν is the specific volume
(ft3/lbm) of the system.
 Entropy is a measure of the inability to do work for a given heat
transferred.
 Power is the time rate of doing work. It is equivalent to the rate of
the energy transfer. Power has units of energy per unit time.
2. Explain the relationship between, work, energy, and power
 Power is defined as the time rate of doing work.
o It is equivalent to the rate of the energy transfer and has
units of energy per unit time.
 Power equivalences:
o 1 𝑓𝑡 _ 𝑙𝑏𝑓/𝑠𝑒𝑐 = 4.6263 𝐵𝑡𝑢/ℎ𝑟 = 1.356 × 10−3 𝑘𝑊
o 1 𝐵𝑡𝑢/ℎ𝑟 = 0.2162 𝑓𝑡 _ 𝑙𝑏𝑓/𝑠𝑒𝑐 = 2.931 × 10−4 𝑘𝑊
o 1 𝑘𝑊 = 3.413 × 103 𝐵𝑡𝑢/ℎ𝑟 = 737.6 𝑓𝑡 _ 𝑙𝑏𝑓/𝑠𝑒𝑐
 Horsepower is related to foot-pounds-force per second (ft-lbf/sec)
by the following relationship:
o 1 ℎ𝑝 = 550.0 𝑓𝑡 _ 𝑙𝑏𝑓/𝑠𝑒𝑐
3. Define the following terms:
 Heat is energy in transit.
 Latent heat is the amount of heat added or removed to produce a
phase change.
 Sensible heat is the heat added or removed that causes a
temperature change.
 Specific heat capacity is the ratio of the heat (Q) added to or
removed from a substance to the resulting change in temperature
(∆T).
 Super heat is the number of degrees a vapor is above the saturation
temperature (boiling point) at a specific pressure.
Objectives
Now that you have completed this lesson, you should be able to do the
following:
1. Define the following thermodynamic properties:
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Rev 1
a. Potential energy
b. Kinetic energy
c. Specific internal energy
d. Specific p-v energy
e. Specific enthalpy
f. Entropy
2. Explain the relationship between work, energy, and power.
3. Define the following:
a. Heat
b. Latent heat
c. Sensible heat
d. Unit used to measure heat
e. Specific heat
f. Super heat
Thermodynamic Units and Properties Summary
In this lesson, new terms were introduced to facilitate the understanding of
thermodynamic properties. Intensive and extensive properties and the
concepts of heat, work, and energy were also introduced. These concepts
and relationships will be used in the remaining chapters on
thermodynamics.
Water, as both a liquid and a vapor, serves as the working fluid in the PWR
energy-conversion cycle. Water absorbs, stores, transports, and transfers
energy throughout the PWR and its supporting systems. Analysis and
evaluation of the entire energy-conversion cycle is conducted by studying
the physical condition of the water at each point of interest within the PWR
and its cycle.
The state of water is defined by two independent intensive thermodynamic
properties. A thermodynamic property is a parameter that describes the
physical condition of a substance. When the state is defined by two
independent intensive properties, all other properties of the state are
implied.
Properties are divided into two classifications, extensive and intensive.
Extensive properties are mass dependent and include mass and volume.
Intensive properties are independent of mass and include temperature,
pressure, and specific volume. Temperature is a measure of the average
kinetic energy of the atoms or molecules of a substance and is measured by
the Fahrenheit and Celsius scales.
The Rankine or Kelvin scales are used for absolute temperature. Pressure is
the force per unit area acting on (or created by) a fluid and is measured in
psig, psiv, psid, inches of Hg (absolute), and inches of Hg (vacuum).
Absolute pressure is given in units of psia. Specific volume is the amount
of space a unit mass occupies and is measured in cubic feet per pound mass.
The inverse of specific volume is density.
Rev 1
43
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Demonstrate understanding of thermodynamic properties, and
methods of measuring intensive and extensive properties.
2. Explain the concepts of heat, work, and energy.
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Rev 1